Determine from the exact solutions for the Blasius and suction boundary layer (given above) what the solution for should be for small and large values of , respectively, and check if these two (asymptotic) results are in agreement with the character of the differential equation given here.

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PROBLEM 1.E – Deformation and vorticity

Each of the expressions given below represents a two-dimensional incompressible flow field, with velocity components u and v in resp. the x- and y-direction of a Cartesian coordinate frame:
(1) u = Ax v = -Ay
(2) u = Ay v = Ax

Here A is a constant, with dimension of [velocity/length]. Assume that A has a positive value.

a. Show that both velocity fields are irrotational and satisfy the (incompressible) continuity equation. Determine for each flow field the shape of the streamlines en sketch the flow pattern (indicate the flow direction).

b. Determine for each of these flow fields how a square fluid element is transported and deformed by the flow over a small time interval Δt. Make a clear sketch for each case. Determine also for each case which (viscous) normal and tangential stresses work on the sides of the fluid element. Indicate these stresses in the sketch, noting the proper direction of the stresses and their relative magnitude.

c. Show that the flow field of (2) is identical to that of (1) and can be obtained by a rotation of the coordinate frame over an angle of 45o. Compare the stress situations that have been derived for each case above, and use this to determine the consequence which the orientation of the fluid element with respect to the flow field has on the stress situation.

d. As the velocity field is irrotational (potential flow), it satisfies the flow equations for both inviscid and viscous flow, and the pressure can be obtained from Bernoulli’s relation (see White, section 2-10). – Compute the pressure field p(x,y) and determine the shape of the isobars in the flow. – Compute explicitly the components of the gradient of the viscous stress tensor ; use this result to explain how the “inviscid potential flow” can also satisfy the viscous flow equations, even though the viscous stresses are not zero.

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PROBLEM 2.C – Stagnation flow on an infinite swept wing
For the flow near the stagnation line of an infinite swept wing the numerical solution is given (see table below) in the form of the nondimensional velocity profiles for the directions respectively perpendicular and parallel to the stagnation line, with:
a. Calculate the velocity profiles us/use and un/use, where the velocity vector has been decomposed w.r.t. the flow direction just outside the boundary layer, for the case that we/ue = 0.5. Scale the components with the outer flow velocity use:

Calculate in addition for this case the cross-flow angle .
Give a graphical representation of the results as profiles of η; also, plot the velocity distributions in the form of a hodograph, i.e. us versus un.

b. Derive the general expression for the cross-flow angle at the wall, , as function of the ratio we/ue. Determine at which we/ue the maximum value of occurs, and what its value is.
Numerical solution:
η u/ue w/we η u/ue w/we η u/ue w/we
0.0 0.00000 0.00000 0.1 0.11826 0.05704 0.2 0.22661 0.11405 0.3 0.32524 0.17091 0.4 0.41446 0.22749 0.5 0.49465 0.28356 0.6 0.56628 0.33889 0.7 0.62986 0.39319 0.8 0.68594 0.44616 0.9 0.73508 0.49751 1.0 0.77786 0.54692
1.1 0.81487 0.59411 1.2 0.84667 0.63883 1.3 0.87381 0.68085 1.4 0.89681 0.72000 1.5 0.91617 0.75616 1.6 0.93235 0.78924 1.7 0.94577 0.81925 1.8 0.95683 0.84619 1.9 0.96588 0.87017 2.0 0.97322 0.89130
2.5 0.99285 0.96058 3.0 0.99842 0.98851 3.5 0.99972 0.99733 4.0 0.99996 0.99951 4.5 1.00000 0.99993
= 1.23259 = 0.57047

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PROBLEM 3.F – Similar expansion boundary layer for β = –1
The Falkner-Skan equation describes the self-similar boundary layer solutions that are obtained for pressure distributions that correspond to an external flow with , and reads:
where and where the following scaling has been used:

In addition to the common boundary conditions, the case of a similar solution with non-zero normal velocity at the wall (suction or injection) has ; where:
For suction vw is negative, hence fw positive. For fw = 0 separation will occur for a certain negative vale of β. A result of the non-linear character of the Falkner-Skan equation is that for negative β not every combination of β and fw allows a solution which possesses a boundary layer character (i.e. with u asymptotically approaching ue), whereas other combinations may display multiple solutions. This is illustrated here for the case β = –1.
a. Determine which (inviscid) expansion geometry corresponds to the case β = –1.
b. Show that for this special case β = –1 the Falkner-Skan equation can be analytically integrated twice. Express the integration constants in the values which characterise the solution at the wall (i.e. and ).
c. Show that, if the solution possesses a boundary layer character, from the behaviour of the solution at large values of η the following relation can be derived:
Hint: use the results of both integration steps! Which values of fw allow boundary layer solutions? Derive that the displacement thickness can be expressed in and .
d. The above shows that for certain values of two different solutions are possible. For one of them is positive and for the other negative (the latter hence shows flow reversal near the wall). Determine both values of when = 2.25 and calculate the two corresponding velocity profiles by a numerical integration (choose any method).
Note: in this case no iteration is required, as all boundary conditions at the wall are known!

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PROBLEM 4.E – Flat plate boundary layer with uniform suction – integral method
The boundary layer on a flat plate with uniform suction (ue(x) = U = constant; vw = constant <0) displays a non-similar development. (see the figure below and section 4-5.2 in White).
Near the leading edge the boundary layer behaves like the Blasius solution, while it develops towards the uniform suction solution for large distance downstream. Properties of these two solutions according to the exact theory are:
1) Blasius solution:
2) Suction profile: where:
Here we apply an integral analysis to compute the (approximate) development of the boundary layer, in particular of the momentum thickness. Under the considered conditions the integral momentum equation is given by:
where: and with:
The shear-parameter S is only a function of the shape of the velocity profile and not of its thickness.
a. The effect of the (constant) parameters U, vw and ν can be conveniently incorporated in a coordinate scaling:
and (remember that !)
– Verify that with these definitions the integral momentum equation can be written as:
with:

– Determine from the exact solutions for the Blasius and suction boundary layer (given above) what the solution for should be for small and large values of , respectively, and check if these two (asymptotic) results are in agreement with the character of the differential equation given here.

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b. The solution of the momentum integral equation requires the value of S, which is not a constant for the entire boundary layer, because the shape of the velocity profile develops with . We may attempt to simplify the problem by using a constant value of S. – Determine from the exact solution data given above the values of S for the Blasius solution and for the suction profile. Integrate the momentum integral equation (numerically) over the domain , for each of these values of S. – Plot the results graphically as and compare it to the accurate results of a finite- difference solution of the problem, given in the table below. Also indicate in the graph the asymptotes of the exact solution as determined in part (a). Comment on the accuracy of the followed approach and the impact of the value of S?
c. As the shape variation has an impact on the computation of the boundary layer, it seems a logical step to try to improve the method by introducing a shape parameter λ, similar as in the method of Thwaites, such that: . Thwaites’method cannot be applied directly, however, as it does not take the effect of suction (both on the shape and in the momentum equation) into account. – The original definition of the shape factor involves the curvature of the velocity profile near the wall:

Evaluate the (differential) x-momentum equation at the wall to show how λ for the general case is related to the pressure gradient and the wall suction velocity. Verify that it reduces to the familiar expression of Thwaites when vw = 0. Derive that for the present problem (where ue(x) = U = constant) it can be directly related to S and . Determine the values of λ for the Blasius and the suction profiles. – Assume that there is approximately a linear relation between S and λ: , and determine the constants such that the Blasius and suction-profile values are reproduced. Show that for the current problem the result can be written as: . – Repeat the numerical integration, now with this modified (adaptive) relation for S. Compare the outcome with the previous results and comment if it matches your expectations.
Table: Numerical results for the flat plate boundary layer with uniform suction. ξ S 0 0.1 0.2 0.3 0.4 0.5 0.7 1.0 1.4 2.0 3.0 4.0 5.0 0 0.1747 0.2293 0.2657 0.2930 0.3147 0.3475 0.3816 0.4113 0.4395 0.4652 0.4787 0.4864 0.2205 0.3008 0.3290 0.3487 0.3640 0.3766 0.3961 0.4172 0.4364 0.4555 0.4736 0.4835 0.4893
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PROBLEM 5.A – Heat transfer from an isothermal flat plate
Consider the development of the thermal boundary layer due to the heat transfer from a heated wall (compressibility and viscous dissipation are neglected). For a flat-plate flow (i.e. constant velocity U in the external flow) over an isothermal surface (i.e. constant wall temperature Tw) a self-similar thermal boundary layer results, where the shape of the temperature profile still depends on the value of the Prandtl number Pr.
The integral temperature equation, which reads as follows for the problem under consideration:
can be used for an approximate estimation of the thermal boundary layer development, when assumed shapes of the velocity and temperature profiles are prescribed. Let now both profiles be approximated by a linear relation, i.e.:
and
and
For the case that δT  δ it has been shown during the lectures that:
where s is the Reynolds-analogy factor. The first expression directly reveals that this situation corresponds to the case that Pr  1.
a. Derive the corresponding expressions for ζ and s for the case that δT  δ. Show that this corresponds to Pr  1. Hints: split the integral in two parts for 0  y  δ and for δ  y  δT. Eliminate δT by introducing ζ (= constant!); δ follows from the integral momentum equation.
b. Investigate the behaviour of both expressions for the limit when Pr goes to zero. N.B.: by “behaviour” it is meant not just the limiting values, but in particular the way in which ζ and s are related to Pr.
c. Give the results of the complete approximation, i.e. for Pr  1 and Pr  1 together, in the form of a graph where 1/ζ and 1/s have been plotted against Pr, on the interval 0  Pr  5. Show that for the limit of Pr going to 1 the results of both approximations match smoothly (continuous in both value and first derivative). N.B.: for this it is sufficient to prove that for ζ and dζ/dPr both limits (i.e. for Pr coming from either side of 1) are equal.
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d. For Pr << 1 the exact solution for the temperature profile can be obtained analytically. As in this case δ << δT, it can be assumed in approximation that the development of the thermal boundary layer takes place completely in the inviscid outer flow. This allows the temperature equation to be simplified to: 2 2Pr TT U x y      – Derive, by means of a similarity transformation, that the above equation can be written as follows: ” 2 ‘ 0      with Θ the transformed temperature profile and η the transformed y-coordinate. Give the expression for η and determine the boundary conditions which θ has to satisfy. N.B.: this η-scaling is not the usual Blasius scaling, which applies to the velocity profile! – Show that the solution of this equation can be written in terms of the Gaussian error-function (see remark below), and derive from this the heat transfer at the wall, expressed by the local Nusselt number: Nu () w x w qx k T T   as function of Rex and Pr. – Calculate from this the expression for the Reynolds analogy factor s, if given that the wall shear stress according to the Blasius solution is: 0.332w U U x    Compare this result to the approximation obtained in part (b). Remark: the error function is: 2 0 2 () x t erf x e dt    , with ( ) 1 erf . (1) VISCOUS FLOWS – AE 4120 8 PROBLEM 6.B – Instability and transition estimates of laminar boundary layers A rough estimate of the location of the point of instability (critical point) or transition can be obtained from semi-empirical correlations, such as e.g.: – for the point of instability: ,Re exp(26.3 8 ) crit H   (after Wieghardt) – for the point of transition: 0.4 ,, Re 2.9(Re ) trans x trans   (after Michel) a. Apply these correlations to determine the point of instability and the point of transition for the following self-similar boundary layer flows. Give in both cases the values for Rex as well as Reθ. -i- flat plate flow: 1/2 2.591 0.664 Rex Hx    -ii- stagnation point flow: 1/2 2.216 0.292 Rex Hx    b. Based on these estimates, what can you conclude about the effect of the pressure gradient on the stability of a laminar boundary layer? c. The velocity profile of the asymptotic suction boundary layer is given by the exponential function: 1 with: w e yvu e u        where w v is the normal velocity at the wall (for suction w v is negative). Determine with the given correlation for the instability point, how large the suction velocity wv must be chosen, in order to keep the boundary layer on the margin of stability. (1) VISCOUS FLOWS – AE 4120 9 PROBLEM 7.C – The Clauser plot The so-called Clauser-plot technique is used to determine the wall shear stress in a turbulent boundary layer from the mean velocity profile, in conditions where an accurate, direct determination of the velocity gradient at the wall is not possible (which is the common situation for turbulent boundary layers). The table given below represents the measured velocity profile, such as determined e.g. by means of a traverse with a pitot tube or a hot-wire probe. As a function of the distance y from the wall (measured in mm), the velocity is given in non-dimensional form as u/Ue, where u is the local (mean) velocity in the boundary layer, and Ue the velocity in the external flow. a. Plot the velocity profile on a semi-logarithmic scale, with u/Ue versus Re / ye yU   . b. Determine the value of the skin friction coefficient Cf by means of a curve fit of the law of the wall to (the lower part of) the velocity profile. c. Determine also the strength of the ‘wake-component’ of the velocity profile. For a measured profile, this is defined as the maximum difference between u(y) and the law of the wall expression. Further data:  = 15.0*10-6 m2/s Ue = 9.804 m/s y(mm) U/Ue 1 0.25 0.219 2 0.50 0.351 3 1.00 0.479 4 1.50 0.535 5 2.00 0.565 6 2.50 0.589 7 3.00 0.607 8 3.50 0.621 9 4.00 0.633 10 4.50 0.644 11 5.00 0.654 12 6.00 0.672 13 7.00 0.687 14 8.00 0.700 15 9.00 0.712 16 10.00 0.723 17 12.00 0.744 18 14.00 0.762 19 16.00 0.779 20 18.00 0.796 y(mm) U/Ue 21 20.00 0.813 22 22.00 0.830 23 24.00 0.847 24 26.00 0.864 25 28.00 0.881 26 30.00 0.899 27 32.00 0.915 28 34.00 0.931 29 36.00 0.946 30 38.00 0.960 31 40.00 0.973 32 42.00 0.983 33 44.00 0.992 34 46.00 0.998 35 48.00 1.000 36 50.00 1.000 37 52.00 1.000 (1) VISCOUS FLOWS – AE 4120 10 PROBLEM 8.A – Turbulence scaling in the wall region + damping functions The wall region of a (2D, incompressible) turbulent boundary layer displays a universal structure, which is known as the law of the wall. According to this concept, the properties of the flow can be expressed uniquely in terms of the so-called wall scaling (also referred to as the use of ‘wall units’). In addition, it is commonly assumed that across the wall region the total shear stress is approximately constant: constant visc turb w       One element of this law of the wall is that the velocity profile of the mean flow satisfies an expression of the following type: () u f y   ; where: *; with: * / * w u yv u y v v       a. Derive from this wall scaling, by means of an elementary dimensional analysis, the scaling of the following (kinematic) properties of the turbulence: 1. the (specific) turbulent shear stress: / ‘ ‘ turb uv   2. the eddy viscosity: t  3. the mixing-length: mix l In the scaling scale only the kinematic properties that are relevant for the wall region ( * v and  ) are used; in other words: determine for each of the above variables how they are to be non- dimensionalised, such that: dimensionless property = function of y+. b. Show for each of these properties how the corresponding “function of y+” is related to the gradient of the velocity profile of the mean flow, that is given by: ‘( ) df du fy dy dy     c. Consider the behaviour of the expressions found in (b), for the limit y+>>1, which constitutes the overlap region, in which the logarithmic velocity profile holds.
Use the results to show that in this overlap region the original (i.e. non-scaled) kinematic turbulence properties mentioned in (a), are independent of viscosity.
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The effect of the viscosity on the turbulence in the wall region is often described by using so-called ‘damping functions’, which are defined as:
‘effective value’ = damping function * ‘fully turbulent value’
where the ‘fully turbulent value’ corresponds to the behaviour of a variable in the overlap region, where the effect of the viscosity can be neglected.
In correspondence to what has been derived in (b), the damping functions for the eddy-viscosity and for the mixing-length can be derived from the shape of the velocity profile.
d. An accurate description of the velocity profile in the entire wall layer, including the viscous region, is given by Spalding’s implicit expression, which reads:
23 ( ) ( )
1
26 Bu uu y u e e u                  
Show that this expression satisfies the two familiar limits of the law of the wall:
y+<<1: uy   (viscous sublayer) y+>>1: 1 ln( ) u y B    (overlap region)
e. – Derive the expressions for both damping functions (i.e., for the eddy viscosity and for the mixing length), as they follow from Spalding’s law of the wall. – Investigate the behaviour of both damping functions for small values of y+ (i.e. determine the first term of the series expansion for small y+). – Plot both damping functions for the interval 0  y+  100. In the calculations use for the constants in the law of the wall the standard values κ = 0.41 and B = 5.0. Hint: maintain in the derivations u+ as the independent variable, and make use of the relation that: 1 / / du dy dy du  